we are given
y varies directly with x
so, we can write equation as

where
k is a constant
now, we need to find k
we are given
y=3 when x=8
so, we can plug it and find k


now, we can plug it back

(A)
we have got equation as

for finding inverse , we can switch x and y

now, we can solve for y


............Answer
(B)
we have

we can plug x=4 and find y

................Answer